Environmental and Exploration Geophysics I
Magnetic Methods
(part II)
tom.h.wilson
[email protected]
Department of Geology and Geography
West Virginia University
Morgantown, WV
Due Tuesday, Dec. 5th
Since the bedrock is
magnetic, we have no
way of differentiating
between anomalies
produced by bedrock
and those produced by buried storage drums.
Acquisition of gravity data allows us to estimate variations
in bedrock depth across the profile. With this knowledge,
we can directly calculate the contribution of bedrock to the
magnetic field observed across the profile.
With the information on bedrock configuration we can clearly
distinguish between the magnetic anomaly associated with
bedrock and that associated with buried drums at the site.
Work using both forward and inverse
calculations
– iterate and adjust…
How many drums are represented by the
triangular-shaped object you entered into your
model?
Use the magic eye to get the coordinates of the
polygon defining the drums
Plot the corner coordinates for the
triangular shaped object you derived at
1:1 scale and compute the area.
How many drums?
Outline of Drum Cluster
Derived from the magnetics model
10
Area of one drum ~
4 square
feet
Depth
15
1
TotalArea  Base x Height
2
Total Area
N Drums 
Area of one Drum
20
25
What’s wrong
with the format of
this plot?
30
35
180
190
200
210
220
Distance along profile
230
We’ll talk about the last
bullet on the results-to-bediscussed list a little later.
www.sensoft.ca
www.sensoft.ca
The potential is the integral of the
force (F) over a displacement path.
V    Fdr   
From above, we obtain a
basic definition of the
potential (at right) for a unit
positive test pole (mt).
p1 p2
dr
2
r
V
pobject
r
Note that we follow Berger
et al. and assume the
leading constants are ~1.
Thus - H (i.e. F/ptest, the field
intensity) can be easily derived
from the potential simply by taking
the derivative of the potential
dV
p
H 
 2
dr r
Consider the field at a point along the axis of a dipole as noted in
problem 1. The dipole in this case could be a buried well casing.
The field has vector properties however, in this case
vectors are collinear and its easy to determine the net effect.
In terms of the
potential we can write
V
p p

r r
In the case at right, r+ is much
greater than r- , thus in
V
p p

r
r
V
p
r
Thus, the potential near either end of a long dipole behaves like
the potential of an isolated monopole. If we are looking for
abandoned wells, we expect to find anomalies similar to the
gravity anomalies encountered over buried spherical objects.
21
Consider the case where the distance to the center of the dipole
is much greater than the length of the dipole. This allows us to
treat the problem of computing the potential of the dipole at an
arbitrary point as one of scalar summation since the directions
to each pole fall nearly along parallel lines.
If r is much much greater than l (distance between the
poles) then the angle  between r+ and r- approaches
0 and r, r+ and r- can be considered parallel so that the
differences in lengths r+ and r- from r equal to plus or
minus the projections of l/2 into r.
r-
r
r+
Working with the potentials of both poles ..
Vdipole 
p p

r
r
Recognizing that pole strength of the
negative pole is the negative of the positive
pole and that both have the same absolute
value, we rewrite the above as
Vdipole  
p p

r r
Vdipole  
p
r  l cos 
2

p
r  l cos 
2
Converting to common denominator yields
Vdipole
pl cos 

r2
where pl = M – the
magnetic moment
From the previous discussion , the field intensity H is just
dV

dr
since V    Fdr, F  
dV
dr
H - monopole =
d  p p

   2
dr
dr  r  r
H - dipole
dVd
d  pl cos  

 

dr
dr  r 2 
dVp
2 pl cos 

r3
This yields the field intensity in the radial direction i.e. in the direction toward the center of the dipole
(along r). However, we can also evaluate the
horizontal and vertical components of the total field
directly from the potential.
H
Toward dipole
(Earth’s) center
ZE  
dVd
d  pl cos  
 

dr
dr  r 2 
Vd represents the
potential of the dipole.
HE is represented by the negative derivative of the
potential along the earth’s surface or in the S direction.
d  pl cos  



rd  r 2 
dV
M sin 

 HE 
3
ds
r
Where M = pl
and
dV 2M cos
ZE  

dr
r3
Let’s tie these results back into some
observations made earlier in the semester
with regard to terrain conductivity data.
32
Given
M sin 
HE 
3
r
What is HE at the equator? … first what’s ?
 is the angle formed by the line connecting the
observation point with the dipole axis. So , in
this case, is a colatitude or 90o minus the
latitude. Latitude at the equator is 0 so  is 90o
and sin (90) is 1.
M
HE  3
r
At the poles,  is 0, so that
HE  0
What is ZE at the equator?
2M cos
ZE 
r3
 is 90
ZE  0
2M cos
ZE 
3
r
ZE at the poles ….
2M
ZE  3
r
The variation of the field intensity at the pole and
along the equator of the dipole may remind you of
the different penetration depths obtained by the
terrain conductivity meters when operated in the
vertical and horizontal dipole modes.
Also compare the field of the
magnetic dipole field to that of the
gravitational monopole field
1
Monopole f ield varies as  2
r
Gravity:500, 1000, 2000m
2M cos
ZE 
3
r
0.12
2M
ZE  3
r
0.1
0.08
0.06
0.04
0.02
0
-1500
-1000
-500
0
500
1000
1500
Increase r by a factor of 4
reduces g by a factor of 16
For the dipole field, an
increase in depth (r) from 4
meters to 16 meters
produces a 64 fold decrease
in anomaly magnitude
1
Dipole field varies as  3
r
Thus the 7.2 nT anomaly (below left) produced by an object at 4
meter depths disappears into the background noise at 16 meters.
0.113 nT
7.2 nT
8
0.15
7
Intensity (nT)
Intensity (nT)
6
5
4
3
2
0.1
0.05
1
0
-1
-5
-3
-1
1
Distance in m eters
3
5
0
-10
-5
0
Distance in m eters
5
10
p
H 2
r
With H in units of Oersteds, p in ups, and r in
cm, we have units of -
ups
1 Oersted  1
2
cm
And the equivalent units for ups
ups  Oersteds-cm
2
5
1 Oersted also equals 10 nT .
Recall you are 20 centimeters from the negative and
positive poles of a dipole as shown below, and that each
pole has pole strength of 1 ups.
Observation
point
20cm
-
+
What is the magnetic field intensity at the observation point
in nanoTeslas?
3. What is the horizontal gradient in nT/m of the Earth’s
vertical field (ZE) in an area where the horizontal field (HE)
equals 20,000 nT and the Earth’s radius is 6.3 x 108 cm.
Recall that horizontal gradients are refer
to the derivative evaluated along the
surface or horizontal direction and we use
the form of the derivative discussed earlier
1 d
r d
dV
dV
d  pl cos  
HE  




ds
rd
rd  r 2 
M sin 
Thus H E 
r3
dV 2M cos
ZE  

3
dr
r
To answer this problem we must evaluate -
1 d
ZE
r d
or
1 d 2M cos 
3
r d
r
Take a minute and give it a try.
4. A buried stone wall constructed from volcanic rocks has
a susceptibility contrast of 0.001cgs emu with its enclosing
sediments. The main field intensity at the site is 55,000nT.
Determine the wall's detectability with a typical proton
precession magnetometer. Assume the magnetic field
produced by the wall can be approximated by a vertically
polarized horizontal cylinder. Refer to figure below, and see
today’s handout.
Background noise at
the site is roughly
5nT.
Vertically Polarized Horizontal Cylinder
2R I

z2
2
Z max
Diagnostic position
X at Z/Zmax
9/10
3/4
2/3
1/2
1/3
1/4
0


2
x


1

2
2
2R I 
z 
Z
z 2   x 2 2 
 1 
2 
z  

x/z
0.188
0.31
0.37
0.495
0.61
0.68
1.0
(x/z)-1
Depth Index multiplier
5.32
3.23
2.7
2.02
1.64
1.47
1
2
x
1
2
Z ( x)
z

Z max  x 2  2
1 

z2 

Vertical Magnetic Anomaly
Vertically Polarized Sphere
Z max
8 3
 R kH
3 3
z
Zmax and ZA refer to the
anomalous field, i.e.
the field produced by the
object in consideration
Diagnostic position
X at Z/Zmax
9/10
3/4
2/3
1/2
1/3
1/4
0
4 3
x2
 R kH (2  2 )
z
ZA  3 3
2
x
z
( 2  1)5/ 2
z
x/z
0.19
0.315
0.377
0.5
0.643
0.73
1.41
(x/z)-1
Depth Index multiplier
5.26
3.18
2.65
2
1.56
1.37
0.71
2

x
2  
z 2 
Z A ( x) 1 

Z max
2  x 2 5 / 2
  1
 z2



The notation can be confusing at times. In
the above, consider H = FE= intensity of
earth’s magnetic field at the survey location.
Vertically Polarized Vertical Cylinder
Z max 
R I
2
z2
Diagnostic position
X at Z/Zmax
9/10
3/4
2/3
1/2
1/3
1/4
x/z
0.27
.046
0.56
0.766
1.04
1.23
(x/z)-1
Depth Index multiplier
3.7
2.17
1.79
1.31
0.96
0.81
zkHA
 R 2 zI
Z A  2 2 3/ 2 , or 2 2 3/ 2
(x  z )
(x  z )
ZA 
R 2 I
1
z2
x2
( 2  1)3 / 2
z
ZA
1
 2
Z max
x
( 2  1)3 / 2
z
Unknow n Anom aly
16
Intensity (nT)
14
12
10
8
6
4
2
0
-4
-3
-2
-1
0
1
2
3
4
Distance in m eters
Sphere, Vertical Cylinder;
ZSphere
z = __________
The depth
Diagnostic
positions
Multipliers
Sphere
Multipliers Cylinder
X3/4 =0.9
3.18
2.17
X1/2 =1.55
2
1.31
X1/4 =2.45
1.37
0.81
ZCylinder
Another Unknown Anomaly
5
Intensity (nT)
4
3
2
1
0
-1
-5
-4
-3
-2
-1
0
1
Distance in meters
2
3
4
5
Sphere or cylinder?
Diagnostic
positions
Multipliers
Sphere
X3/4 = 1.6
ZSphere
Multipliers
Cylinder
ZCylinder
3.18
2.17
2.82
X1/2 = 2.5
2
1.31
2.69
X1/4 = 3.7
1.37
0.81
2.34
6. Given that Z max 
 R2 I
derive an expression for the radius,
2
z
where I = kHE. Compute the depth to the top of the casing for
the anomaly shown below, and then estimate the radius of the
casing assuming k = 0.1 and HE =55000nT. Zmax (62.2nT from
graph below) is the maximum vertical component of the
anomalous field produced by the vertical casing.
Abandoned well
70
Intensity (nT)
60
50
40
30
20
10
0
-15
-10
-5
0
Distance in m eters
5
10
15
TODAY
Tuesday Dec. 5th
Thursday Dec. 7th
Thursday Dec. 7th